Justifying assumptions about method to find equilibrium pressure for co-existence of graphite and diamond

Question

The standard state Gibbs free energies of formation of $\ce{C(graphite)}$ and $\ce{C(diamond)}$ at $T = \pu{298 K}$ are $\pu{0 kJ mol-1}$ and $\pu{2.9 kJ mol-1}$, respectively.

The conversion of graphite $\ce{C(graphite)}$ to diamond $\ce{C(diamond)}$ reduces its volume by $\pu{2e-6 m3 mol-1}$.

If $\ce{C(graphite)}$ is converted to $\ce{C(diamond)}$ isothermally at $T = \pu{298 K}$, the pressure at which $\ce{C(graphite)}$ is in equilibrium with $\ce{C(diamond)}$ is:

(A) $\pu{14501 bar}$
(B) $\pu{58001 bar}$
(C) $\pu{1450 bar}$
(D) $\pu{29001 bar}$

JEE Adv, 2017, related 1, 2

In the solution stated in this site, there are two assumptions made about this process.

1. $\Delta S_r = 0$ i.e. the total entropy change of the process is zero.
2. The internal energy change is zero due to it being an isothermal process.
3. $\Delta H_r= P \Delta V$, that is we assume pressure is constant for the process and solve for $P$. The total pressure in the final state is apparently the initial pressure plus the pressure calculate from the ratio $\frac{\Delta H_r}{\Delta V}$.

How do we justify these two assumptions?

Firstly, how can we justify the entropy change being zero? And, the second point I don't get how we can claim $\Delta U = 0$, just because it is isothermal. I know it is true for ideal gases but how does that apply here? About the last assumption, I can not understand at all what the logic is behind finding the final pressure as initial plus the calculated from the ratio.

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P.S: I know the given links answer the question completely but I want to figure out how to reason these assumptions myself.

Answer 1

This is definitely not the approach that I would have used to solve this problem. I would have used the condition that, at the equilibrium pressure $G_g=G_d$ or $\Delta G=0$. The free energy of graphite at the equilibrium condition would be $$G_g=G_g^0+v_g\Delta P$$where $G_g^0$ is the molar free energy of graphite in the standard state (0 kJ/mol), $v_g$ is the molar volume of graphite, and $\Delta P$ is the increase in pressure relative to the standard state. Similarly, for diamond, at the equilibrium condition, $$G_d=G_d^0+(v_g-\Delta v)\Delta P$$where $G_d^0$ is the molar free energy of graphite in the standard state (2.9 kJ/mol), and $\Delta v=2\times 10^{-6}\ m^3/mol$. So setting $\Delta G =0$ gives $$0=(G_d^0-G_g^0)-(\Delta v)(\Delta P)$$